009B Sample Midterm 3, Problem 2

State the fundamental theorem of calculus, and use this theorem to find the derivative of

$F(x)=\int _{\cos(x)}^{5}{\frac {1}{1+u^{10}}}~du.$

Foundations:
What does Part 1 of the Fundamental Theorem of Calculus
say is the derivative of $G(x)=\int _{x}^{5}{\frac {1}{1+u^{10}}}~du?$
First, we need to switch the bounds of integration.
So, we have $G(x)=-\int _{5}^{x}{\frac {1}{1+u^{10}}}~du.$
By Part 1 of the Fundamental Theorem of Calculus, $G'(x)=-{\frac {1}{1+x^{10}}}.$

Solution:

Step 1:
The Fundamental Theorem of Calculus, Part 1
Let $f$ be continuous on $[a,b]$ and let $F(x)=\int _{a}^{x}f(t)~dt.$
Then, $F$ is a differentiable function on $(a,b)$ and $F'(x)=f(x).$
The Fundamental Theorem of Calculus, Part 2
Let $f$ be continuous on $[a,b]$ and let $F$ be any antiderivative of $f.$ Then,
$\int _{a}^{b}f(x)~dx=F(b)-F(a).$

Step 2:
First,
$F(x)=-\int _{5}^{\cos(x)}{\frac {1}{1+u^{10}}}~du.$
Now, let $g(x)=\cos(x)$ and $G(x)=\int _{5}^{x}{\frac {1}{1+u^{10}}}~du.$
Therefore,
$F(x)=-G(g(x)).$
Hence,
$F'(x)=-G'(g(x))g'(x)$
by the Chain Rule.

Step 3:
Now,
$g'(x)=-\sin(x).$
By the Fundamental Theorem of Calculus,
$G'(x)={\frac {1}{1+x^{10}}}.$
Hence,
${\begin{array}{rcl}\displaystyle {F'(x)}&=&\displaystyle {-{\frac {1}{1+\cos ^{10}x}}(-\sin(x))}\\&&\\&=&\displaystyle {{\frac {\sin(x)}{1+\cos ^{10}x}}.}\\\end{array}}$

Final Answer:
The Fundamental Theorem of Calculus, Part 1
Let $f$ be continuous on $[a,b]$ and let $F(x)=\int _{a}^{x}f(t)~dt.$
Then, $F$ is a differentiable function on $(a,b)$ and $F'(x)=f(x).$
The Fundamental Theorem of Calculus, Part 2
Let $f$ be continuous on $[a,b]$ and let $F$ be any antiderivative of $f.$ Then,
$\int _{a}^{b}f(x)~dx=F(b)-F(a).$
$F'(x)={\frac {\sin(x)}{1+\cos ^{10}x}}$

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